[leetcode-763] Partition Labels

最新推荐文章于 2022-04-04 16:12:37 发布

原创最新推荐文章于 2022-04-04 16:12:37 发布 · 206 阅读

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本文介绍了一种高效的算法，用于将一个由小写英文字母组成的字符串分割成尽可能多的部分，使得每个字母只出现在一个部分中。通过查找每个字符的首次和最后一次出现的位置，该算法能够确定每个部分的边界，并返回表示各部分长度的整数列表。

Description

A string S of lowercase English letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.

Solution

Select a partition with the first and last occurrence of a letter, mark them as start and end of a list respectively.
Iterate all letters within the list.
If found letter have last occurrence bigger than end, update the end accordingly.
If end was reached, update count with end-start+1
Start new round with end + 1

code:

class Solution {
public:
    vector<int> partitionLabels(string S) {
        vector<int> result;
        int lastOccurence[26];
        
        for(int i = 0; i < S.length(); i++) {
            lastOccurence[S[i] - 'a'] = i;
        }
        
        int i = 0;
        while(i < S.length()) {
            int end = lastOccurence[S[i] - 'a'];
            
            int j = i;
            
            while(j<end) {
                end = max(end, lastOccurence[S[j] - 'a']);
                j++;
            }
            
            int count = j - i + 1;
            result.push_back(count);
            
            i = j + 1;
        }
        
        return result;
        
    }
};